Are All Triangles Isosceles?

Date: 09/15/1999 at 13:03:37
From: D. Klotz
Subject: Are all triangles isosceles?
There is a rather involved (and it is claimed, well-known) proof that
shows that all triangles are isosceles (it can be found in Euclidean
and non-Euclidean geometries - Marvin Jay Greenberg, bottom of pg. 23)
- but unfortunately after studying it, I cannot seem to find the flaw
in the argument. Your help would be much appreciated. It claims:
Given triangle ABC. Construct the bisector of angle A and the
perpendicular bisector of side BC opposite to angle A. Now consider
the various cases (there are diagrams given in the book).
Case 1: The bisector of angle A and the perpendicular bisector of
segment BC are either parallel or identical. In either case, the
bisector of angle A is perpendicular to BC and hence, by definition,
is an altitude. Therefore the triangle is isosceles (The conclusion
follows from the Euclidean theorem that states: if an angle bisector
and altitude from the same vertex of a triangle coincide, the triangle
is isosceles.)
Suppose now that the bisector of angle A and the perpendicular
bisector of the side opposite are not parallel and do not coincide.
Then they intersect in exactly one point, D, and there are 3 cases to
consider:
Case 2: The point D is inside the triangle
Case 3: The point D is on the triangle
Case 4: The point D is outside the triangle
For each case, construct DE perpendicular to AB and DF perpendicular
to AC, and for cases 2 and 4 join D to B and D to C. In each case the
following proof now holds:
(I don't have the appropriate symbol for congruence on my keyboard so
I'll use '=' to mean congruence.)
DE = DF because all points on an angle bisector are equidistant from
the sides of the angle
DA = DA, and angle DEA and angle DFA are right angles
Hence triangle ADE is congruent to triangle ADF by the hypotenuse-leg
theorem of Euclidean Geometry. Therefore, we have AE = AF.
Now, DB = DC because all points on the perpendicular bisector of a
segment are equidistant from the ends of the segment.
Also, DE = DF, and angle DEB and angle DFC are right angles.
Hence, triangle DEB is congruent to triangle DFC by the hypotenuse-leg
theorem, and hence FC = BE.
It follows that AB = AC, in cases 2 and 3 by addition, and in case 4
by subtraction. The triangle is therefore isosceles.
QED

Date: 09/15/1999 at 14:00:47
From: Doctor Rob
Subject: Re: Are all triangles isosceles??
Thanks for writing to Ask Dr. Math, D.
This is one of the classic fallacious proofs. The error is rather
subtle. First of all, cases 2 and 3 are impossible. Secondly, in case
4, either E is between A and B but F is not between A and C, or
vice-versa. Thus you get that
AB = AE + EB = AF + FC, but AC = AF - FC
or else
AB = AE - EB = AF - FC, but AC = AF + FC
These cannot be equal unless both FC and EB are zero, which cannot
happen since we are not in Case 1.
If you draw a diagram carefully, using accurate instruments, you will
see that what I say is true. Or, you could prove it.
- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/